Absolute Convergence and Divergence — Question 6

*Question 6 Consider $\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^n}{n\ln n}$.

1. Verify ordinary convergence with the AST.

2. Test absolute convergence using the logarithmic benchmark.

3. Classify the series and state an alternating error estimate.

*Question 6 – Solution

Let $b_n=1/(n\ln n)$. For $n\ge2$ it is positive. Its denominator is increasing, so $b_n$ decreases, and $b_n\to0$. Thus the alternating series converges.

The absolute series is $\sum1/(n\ln n)$. For $f(x)=1/(x\ln x)$, $$\int_2^b f(x)\,dx=\ln(\ln b)-\ln(\ln2)\to\infty.$$ The Integral Test proves divergence.

The original series is conditionally convergent, with $$|S-s_N|\le\frac1{(N+1)\ln(N+1)}.$$